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In Vasicek one-factor model (and in other affine models), the price of a zero-coupon bond at time $t$ conditional on the information at this time is

$$P(t,T) = E[e^{-\int^T_tr(u)du}|F_t] = A(t,T)e^{-B(t,T)r(t)} \quad\quad (1)$$ for which $r(t)$ is known since we have information $F_t$ and so this price $P(t,T)$ is deterministic at time $t$ (if I understand it correctly).

But what if we only have information $F_s$ with $s<t$, what is the price of this bond at time $t$ "as seen from time $s$" (which now should be a random variable rather than deterministic)? Do we "just replace" $r(t)$ above with the solution to Vasicek model, i.e. with $$r(t) = r(s)e^{-k(t-s)} + \theta(1-e^{-k(t-s)}) + \sigma\int^t_se^{-k(t-u)}dW(u) ?\quad\quad (2)$$

Is yes, then what is the formal justification for it (in terms of the pricing formula via conditional expectation)?

Putting this another way: what is the SDE for zero-coupon bonds in Vasicek model?

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I think the use of the same variable $t$ in (1) and (2) causes some confusion (I know it does for me) so let me re-write $(1)$ as $(1^*)$

$$P(s,T) = E[e^{-\int^T_s r(u)du}|F_s] = A(s,T)e^{-B(s,T)r(s)} \quad\quad (1^*)$$

which is the price of the bond at time $s$ given the information $F_s$ up to this time.

What is the price of the bond at time $t>s$ "as seen from" $s$ given the information $F_s$? How is it expressed in terms of the conditional expectation?

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  • $\begingroup$ I think you are referring to forward bond prices. Generally speaking the price of a forward bond is given by $P(s,t,T)=P(s,T)/P(s,t)$. Indeed, rearranging: $P(s,t)P(s,t,T)=P(s,T)$, meaning that the forward price must be so that it is equivalent 1) to invest your capital at $s$ up to $T$, and 2) to invest your capital at $s$ up to $t$, then to invest it again from $t$ up to $T$ at a ZC rate set at initial time $s$. $\endgroup$ – Daneel Olivaw Jun 5 '18 at 17:59
  • $\begingroup$ @DaneelOlivaw Thank you for your reply. I am not sure, however, if it is what I was looking for, since your formula gives a deterministic quantity at time $s$: we have information $F_s$ up to this time and hence both $P(s,T)$ and $P(s,t)$ in your equation are deterministic. $\endgroup$ – Confounded Jun 5 '18 at 18:05
  • $\begingroup$ Ok, with your modified question I see what you mean. Note that, because $r(t)$ is normal and $A(t,T)$ a deterministic quantity, your zero-coupon bond is log-normal. You can derive its SDE, with drift and diffusion parameters that depend on $k$, $\theta$ and $\sigma$, and obtain an expression for $P(t,T)$ that depends on $P(s,T)$ similar to the solution of the stock price process under the Black-Scholes model. $\endgroup$ – Daneel Olivaw Jun 5 '18 at 19:32
  • $\begingroup$ @DaneelOlivaw "because $r(t)$ is normal" - but $r(t)$ in $P(t,T)$ in (1) (which is calculated as the expectation conditional on $F_t$) is not normal, it is deterministic, i.e. just a number. $r(t)$ is normal in (2). $\endgroup$ – Confounded Jun 5 '18 at 19:42
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    $\begingroup$ If you have access to Shreve's book Stochastic Calculus for Finance II check theorem 6.3.1, this should help clarify concepts. Your expectation is saying "if $r(t) = \text{some value}$ then $P(t,T)=\dots$", and it turns out you can replace this $\text{some value}$ $-$ i.e. some realization of $r(t)$ $-$ by the random variable $r(t)$ itself. $\endgroup$ – Daneel Olivaw Jun 5 '18 at 20:10

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